Question

Evaluate each of the following using identities:

(i) $(399{)}^2$

(ii) $(0.98{)}^2$

(iii) $991\times 1009$

(iv) $117\times 83$

Answer 1

$\left(i\right)(399{)}^2=(400-1{)}^2=(400{)}^2-2\times 400\times 1+(1{)}^2\{\because (a-b{)}^2=a^2-2ab+b^2\}=160000-800+1=160001-800=159201(ii\left)\right(0.98{)}^2=(1-0.02{)}^2\{\because (a-b{)}^2=a^2-2ab+b^2\}=(1{)}^2-2\times 1\times 0.02+(0.02{)}^2=1-0.04+0.0004=1.0004-0.04=1.0004-0.0400=0.9604\left(iii\right)991\times 1009=(1000-9)(1000+9)\{\because (a+b\left)\right(a-b)=a^2-b^2\}=(1000{)}^2-(9{)}^2=1000000-81=999919\left(iv\right)117\times 83=(100+17)(100-17)\{\because (a+b\left)\right(a-b)=a^2-b^2\}=(100{)}^2-(17{)}^2=10000-289=9711$